# Number of elements of order $p$ in a (possibly infinite) group

If a group is finite, then it's easy to see by group action that $$n_p\equiv-1\pmod p$$.

However, is the result true for an infinite group? My conjecture is that if $$n_p$$ is non-zero and finite, then $$n_p\equiv-1\pmod p$$ of infinite groups. I proved the result for $$n_2$$, but don't know whether it is true for all primes.

• To be clear, $n_p$ is the number of elements of order $p$? Jan 9, 2021 at 19:21
• @HallaSurvivor yes Jan 9, 2021 at 19:25
• The deleted answer of @HallaSurvivor is actually correct. If there are only finitely many elements of order $p$ then they generate a finite subgroup $H$. $C_p*C_p$ is not a counterexample, because it has infinitely many elements of order $p$. To see that $H$ is finite, note that its centre has finite index, and so (by a result of Schur) $[H,H]$ is finite, but $H/[H,H]$ is also finite. Jan 9, 2021 at 19:49
• @DerekHolt -- I've made my answer community wiki. Would you mind elaborating on your comment slightly in an edit? I'm not sure I follow. Jan 9, 2021 at 19:57

We know that the result is true for finite groups (by the group-action argument you're familiar with).

But if there's only finitely many elements of order $$p$$, we can look at $$H$$ the group generated by those elements. It is finite, and contains all the elements of order $$p$$. So the result holds for $$H$$, and since $$H$$ has all the elements of order $$p$$, the result holds for $$G$$.

I hope this helps ^_^

Added by Derek Holt: proof that $$H$$ is finite. Since $$H$$ is generated by the elements of order $$p$$ and there are only finitely many, they have only finitely many conjugates, so their centralizers have finite index in $$H$$. So the intersection of their centralizers, which is the centre $$Z(H)$$ of $$H$$ has finite index in $$H$$.

Now, by a result of Schur, the derived group $$[H,H]$$ is finite. Since $$H/[H,H]$$ is an abelian group generated by finitely many elements of order $$p$$, it is also finite, and hence $$H$$ is finite.

Suppose $$G$$ is a group with finitely many elements of order $$p$$. We may assume that $$G$$ is generated by its elements of order $$p$$ (otherwise, just look at the subgroup they generate). Since every conjugate of an element of order $$p$$ has order $$p$$, each element of order $$p$$ must have centralizer of finite index. So, the intersection $$Z$$ of the centralizers of elements of order $$p$$ is finite index. Since $$G$$ is generated by its elements of order $$p$$, $$Z$$ is actually the center of $$G$$.

So, the center of $$G$$ has finite index. This implies that the commutator subgroup $$[G,G]$$ is finite (see this question for instance). But $$G/[G,G]$$ is an abelian group generated by finitely many elements of finite order, so it is finite as well. Thus $$G$$ is finite, and so the result for finite groups applies.

• Side remark: The argument works equally when the set of elements of order $p$ is replaced with any finite conjugacy-invariant subset $S$ consisting of elements of finite order, to show that $\langle S\rangle$ is finite.
– YCor
Jan 10, 2021 at 23:05